Find particular solution of $$u''+2u'+u=9x^2e^{-x}.$$ So I've already worked out that the particular solution is $Ae^{-x}+Bxe^{-x}$ so then I tried the particular solution $u_p=Cx^2e^{-x}$ but it didn't work.
Anybody know why it didn't work and what I should try instead?
On
Robert's hint is the best.Here is another approach $$u′′ + 2u′ + u = 9x^2e^{−x}$$ $$(u''e^x+u'e^x+u'e^x+ue^x=9x^2$$ $$(u'e^x)'+(ue^x)'=9x^2$$ $$u'e^x+ue^x=3x^3+K$$ $$(ue^x)'=3x^3+K$$ $$ue^x=\frac 34 x^4+Kx+C$$ $$ \implies u(x)=e^ {-x}\left(\frac 34 x^4+Kx+C\right)$$
On
A very simple approach to handle such questions is using differential operators (due to Heaviside)
$$(D+1)^2u = 9x^2e^{-x}$$
Now using exponential shift, we have: $$u_p = e^{-x}\frac{1}{D^2} 9x^2 $$
which is integrating $x^2$ twice.
Refer this document for operator methods of ODE : math.mit.edu
On
$u''+2u'+u=9x^2e^{-x}\implies (D^2+2D'+1)u=9x^2e^{-x}$
Particular Integral (P.I.) $\quad = \frac{1}{D^2+2D'+1}(9x^2e^{-x})$
$= 9\frac{1}{(D+1)^2}x^2e^{-x}$
$= 9e^{-x} \frac{1}{\{(D-1)+1\}^2}x^2$
$= 9e^{-x} \frac{1}{D^2}x^2$
$=\frac{3}{4}x^4e^{-x}$
For the Particular Integral (i.e., P.I.) there are some general rules
$1.$ $\frac{1}{D + a} \phi (x) = e^{-ax}\int e^{ax}\phi(x)$
$2.$ $\frac{1}{f(D)} e^{ax} \phi(x) = e^{ax}\frac{1}{f(D+a)} \phi(x)$
$3.$ $\frac{1}{f(D)} x^{n} \sin ax = $Imaginary part of $e^{iax}\frac{1}{f(D+ia)} x^n$
$4.$ $\frac{1}{f(D)} x^{n} \cos ax = $Real part of $e^{iax}\frac{1}{f(D+ia)} x^n$
$5.$ $\frac{1}{f(D)} x^{n} (\cos ax + i\sin ax) = \frac{1}{f(D)} x^n e^{iax}=e^{iax}\frac{1}{f(D+ia)} x^n$
Since the polynomial on the right-hand side is of second degree and $-1$ is a root of the characteristic polynomial with multiplicity $2$, you should try the following more general form $$u_p(x)=x^2(Ax^2+Bx+C)e^{-x}.$$ More generally if the right-hand side is $e^{rx}P(x)$ where $P$ is a polynomial of degree $d$ and $r$ is a root of the characteristic polynomial with multiplicity $m$ then use the form $$u_p(x)=x^mQ(x)e^{rx}$$ where $Q$ is a generic polynomial of degree $d$.